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Isolated Singularities, and the Residue Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: this module covers the residue theorem, removable and isolated singularities, some new developments from Cauchy's theorem, and related exercises.

The first result we present in this section is a natural extension of (Reference). However, as we shall see, its consequences for computing contour integrals can hardly be overstated.

Theorem 1

Let SS be a piecewise smooth geometric set whose boundary CSCS has finite length. Suppose c1,...,cnc1,...,cn are distinct points in the interior S0S0 of S,S, and that r1,...,rnr1,...,rn are positive numbers such that the closed disks {B¯rk(ck)}{B¯rk(ck)} are contained in S0S0 and pairwise disjoint. Suppose ff is continuous on S∖∪Brk(ck),S∖∪Brk(ck), i.e., at each point of SS that is not in any of the open disks Brk(ck),Brk(ck), and that ff is differentiable on S0∖∪B¯rk(ck),S0∖∪B¯rk(ck), i.e., at each point of S0S0 that is not in any of the closed disks B¯rk(ck).B¯rk(ck). Write CkCk for the circle that is the boundary of the closed disk B¯rk(ck).B¯rk(ck). Then

∫ C S f ( ζ ) d ζ = ∑ k = 1 n ∫ C k f ( ζ ) d ζ . ∫ C S f ( ζ ) d ζ = ∑ k = 1 n ∫ C k f ( ζ ) d ζ .
(1)

Proof

This is just a special case of part (d) of (Reference).

Let ff be continuous on the punctured disk B'¯r(c),B'¯r(c), analytic at each point zz in Br'(c),Br'(c), and suppose ff is undefined at the central point c.c. Such points cc are called isolated singularities of f,f, and we wish now to classify these kinds of points. Here is the first kind:

Definition 1:

A complex number cc is called a removable singularity of an analytic function ff if there exists an r>0r>0 such that ff is continuous on the punctured disk B'¯r(c),B'¯r(c), analytic at each point in Br'(c),Br'(c), and limz→cf(z)limz→cf(z) exists.

Exercise 1

  1. Define f(z)=sinz/zf(z)=sinz/z for all z≠0.z≠0. Show that 0 is a removable singularity of f.f.
  2. For z≠c,z≠c, define f(z)=(1-cos(z-c))/(z-c).f(z)=(1-cos(z-c))/(z-c). Show that cc is a removable singularity of f.f.
  3. For z≠c,z≠c, define f(z)=(1-cos(z-c))/(z-c)2.f(z)=(1-cos(z-c))/(z-c)2. Show that cc is still a removable singularity of f.f.
  4. Let gg be an analytic function on Br(c),Br(c), and set f(z)=(g(z)-g(c))/(z-c)f(z)=(g(z)-g(c))/(z-c) for all z∈Br'(c).z∈Br'(c). Show that cc is a removable singularity of f.f.

The following theorem provides a good explanation for the term “removable singularity.” The idea is that this is not a “true” singularity; it's just that for some reason the natural definition of ff at cc has not yet been made.

Theorem 2

Let ff be continuous on the punctured disk B¯r'(c)B¯r'(c) and differentiable at each point of the open punctured disk Br'(c),Br'(c), and assume that cc is a removable singularity of f.f. Define f˜f˜ by f˜(z)=f(z)f˜(z)=f(z) for all z∈Br'(c),z∈Br'(c), and f˜(c)=limz→cf(z).f˜(c)=limz→cf(z). Then

  1.  f˜f˜ is analytic on the entire open disk Br(c),Br(c), whence
    f(z)=∑k=0∞ck(z-c)kf(z)=∑k=0∞ck(z-c)k
    (2)
    for all z∈Br'(c).z∈Br'(c).
  2. For any piecewise smooth geometric set S⊆Br(c),S⊆Br(c), whose boundary CSCS has finite length, and for which c∈S0,c∈S0,
    ∫CSf(ζ)dζ=0.∫CSf(ζ)dζ=0.
    (3)

Proof

As in part (a) of (Reference), define FF on Br(c)Br(c) by

F ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ . F ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ .
(4)

Then, by that exercise, FF is analytic on Br(c).Br(c). We show next that F(z)=f˜(z)F(z)=f˜(z) on Br(c),Br(c), and this will complete the proof of part (1).

Let zz be a point in Br(c)Br(c) that is not equal to c,c, and let ϵ>0ϵ>0 be given. Choose δ>0δ>0 such that δ<|z-c|/2δ<|z-c|/2 and such that |f˜(ζ)-f˜(c)|<ϵ|f˜(ζ)-f˜(c)|<ϵ if |ζ-c|<δ.|ζ-c|<δ. Then, using part (c) of (Reference), we have that

f ˜ ( z ) = f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ - 1 2 π i ∫ C δ f ( ζ ) ζ - z d ζ = F ( z ) - 1 2 π i ∫ C δ f ( ζ ) - f ˜ ( c ) ζ - z d ζ - 1 2 π i ∫ C δ f ˜ ( c ) ζ - z d ζ = F ( z ) - 1 2 π i ∫ C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ , f ˜ ( z ) = f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ - 1 2 π i ∫ C δ f ( ζ ) ζ - z d ζ = F ( z ) - 1 2 π i ∫ C δ f ( ζ ) - f ˜ ( c ) ζ - z d ζ - 1 2 π i ∫ C δ f ˜ ( c ) ζ - z d ζ = F ( z ) - 1 2 π i ∫ C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ ,
(5)

where the last equality holds because the function f˜(c)/(ζ-z)f˜(c)/(ζ-z) is an analytic function of ζζ on the disk Bδ(c),Bδ(c), and hence the integral is 0 by (Reference). So,

| f ˜ ( z ) - F ( z ) | = | 1 2 π i ∫ C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ | ≤ 1 2 π ∫ C δ | f ˜ ( ζ ) - f ˜ ( c ) | | ζ - z | d s ≤ 1 2 π ∫ C δ ϵ δ / 2 d s = 2 ϵ δ × δ = 2 ϵ . | f ˜ ( z ) - F ( z ) | = | 1 2 π i ∫ C δ f ˜ ( ζ ) - f ˜ ( c ) ζ - z d ζ | ≤ 1 2 π ∫ C δ | f ˜ ( ζ ) - f ˜ ( c ) | | ζ - z | d s ≤ 1 2 π ∫ C δ ϵ δ / 2 d s = 2 ϵ δ × δ = 2 ϵ .
(6)

Since this holds for arbitrary ϵ>0,ϵ>0, we see that f˜(z)=F(z)f˜(z)=F(z) for all z≠cz≠c in Br(c).Br(c).

Finally, since

f ˜ ( c ) = lim z → c f ˜ ( z ) = lim z → c F ( z ) = F ( c ) , f ˜ ( c ) = lim z → c f ˜ ( z ) = lim z → c F ( z ) = F ( c ) ,
(7)

the equality of FF and f˜f˜ on all of Br(c)Br(c) is proved. This finishes the proof of part (1).

Exercise 2

Prove part (2) of the preceding theorem.

Now, for the second kind of isolated singularity:

Definition 2:

A complex number cc is called a pole of a function ff if there exists an r>0r>0 such that ff is continuous on the punctured disk B'¯r(c),B'¯r(c), analytic at each point of Br'(c),Br'(c), the point cc is not a removable singularity of f,f, and there exists a positive integer kk such that the analytic function (z-c)kf(z)(z-c)kf(z) has a removable singularity at c.c.

A pole cc of ff is said to be of order n, if nn is the smallest positive integer for which the function f˜(z)≡(z-c)nf(z)f˜(z)≡(z-c)nf(z) has a removable singularity at c.c.

Exercise 3

  1. Let cc be a pole of order nn of a function f,f, and write f˜(z)=(z-c)nf(z).f˜(z)=(z-c)nf(z). Show that f˜f˜ is analytic on some disk Br(c).Br(c).
  2. Define f(z)=sinz/z3f(z)=sinz/z3 for all z≠0.z≠0. Show that 0 is a pole of order 2 of f.f.

Theorem 3

Let ff be continuous on a punctured disk B'¯r(c),B'¯r(c), analytic at each point of Br'(c),Br'(c), and suppose that cc is a pole of order nn of f.f. Then

  1. For all z∈Br'(c),z∈Br'(c),
    f(z)=∑k=-n∞ak(z-c)k.f(z)=∑k=-n∞ak(z-c)k.
    (8)
  2. The infinite series of part (1) converges uniformly on each compact subset KK of Br'(c).Br'(c).
  3. For any piecewise smooth geometric set S⊆Br(c),S⊆Br(c), whose boundary CSCS has finite length, and satisfying c∈S0,c∈S0,
    ∫CSf(ζ)dζ=2πia-1,∫CSf(ζ)dζ=2πia-1,
    (9)
    where A-1A-1 is the coefficient of (z-c)-1(z-c)-1 in the series of part (1).

Proof

For each z∈Br'(c),z∈Br'(c), write f˜(z)=(z-c)nf(z).f˜(z)=(z-c)nf(z). Then, by Theorem 2, f˜f˜ is analytic on Br(c),Br(c), whence

f ( z ) = f ˜ ( z ) ( z - c ) n = 1 ( z - c ) n ∑ k = 0 ∞ c k ( z - c ) k = ∑ k = - n ∞ a k ( z - c ) k , f ( z ) = f ˜ ( z ) ( z - c ) n = 1 ( z - c ) n ∑ k = 0 ∞ c k ( z - c ) k = ∑ k = - n ∞ a k ( z - c ) k ,
(10)

where ak=cn+k.ak=cn+k. This proves part (1).

We leave the proof of the uniform convergence of the series on each compact subset of Br'(c),Br'(c), i.e., the proof of part (2), to the exercises.

Part (3) follows from Cauchy's Theorem ((Reference)) and the computations in (Reference). Thus:

∫ C S f ( ζ ) d ζ = ∫ C r f ( ζ ) d ζ = ∫ C r ∑ k = - n ∞ a k ( z - c ) k d ζ = ∑ k = - n ∞ a k ∫ C r ( ζ - c ) k d ζ = a - 1 2 π i , ∫ C S f ( ζ ) d ζ = ∫ C r f ( ζ ) d ζ = ∫ C r ∑ k = - n ∞ a k ( z - c ) k d ζ = ∑ k = - n ∞ a k ∫ C r ( ζ - c ) k d ζ = a - 1 2 π i ,
(11)

as desired. The summation sign comes out of the integral because of the uniform convergence of the series on the compact circle Cr.Cr.

Exercise 4

  1. Complete the proof to part (2) of the preceding theorem. That is, show that the infinite series ∑k=-n∞ak(z-c)k∑k=-n∞ak(z-c)k converges uniformly on each compact subset KK of Br'(c).Br'(c). HINT: Use the fact that the Taylor series ∑n=0∞cn(z-c)n∑n=0∞cn(z-c)n for f˜f˜ converges uniformly on the entire disk B¯r(c),B¯r(c), and that if cc is not in a compact subset KK of Br(c),Br(c), then there exists a δ>0δ>0 such that |z-c|>δ|z-c|>δ for all z∈K.z∈K.
  2. Let f,c,f,c, and f˜f˜ be as in the preceding proof. Show that
    a-1=f˜(n-1)(c)(n-1)!.a-1=f˜(n-1)(c)(n-1)!.
    (12)
  3. Suppose gg is a function defined on a punctured disk Br'(c)Br'(c) that is given by the formula
    g(z)=∑k=-n∞ak(z-c)kg(z)=∑k=-n∞ak(z-c)k
    (13)
    for some positive integer nn and for all z∈Br'(c).z∈Br'(c). Suppose in addition that the coefficient a-n≠0.a-n≠0. Show that cc is a pole of order nn of g.g.

Having defined two kinds of isolated singularities of a function f,f, the removable ones and the polls of finite order, there remain all the others, which we collect into a third type.

Definition 3:

Let ff be continuous on a punctured disk B'¯r(c),B'¯r(c), and analytic at each point of Br'(c).Br'(c). The point cc is called an essential singularity of ff if it is neither a removable singularity nor a poll of any finite order. Singularities that are either poles or essential singularities are called nonremovable singularities.

Exercise 5

For z≠0,z≠0, define f(z)=e1/z.f(z)=e1/z. Show that 0 is an essential singularity of f.f.

Theorem 4

Let ff be continuous on a punctured disk B'¯r(c),B'¯r(c), analytic at each point of Br'(c),Br'(c), and suppose that cc is an essential singularity of f.f. Then

  1. For all z∈Br'(c),z∈Br'(c),
    f(z)=∑k=-∞∞ak(z-c)k,f(z)=∑k=-∞∞ak(z-c)k,
    (14)
    where the sequence {ak}-∞∞{ak}-∞∞ has the property that for any negative integer NN there is a k<Nk<N such that ak≠0.ak≠0.
  2. The infinite series in part (1) converges uniformly on each compact subset KK of Br'(c).Br'(c). That is, if FnFn is defined by Fn(z)=∑k=-nnak(z-c)k,Fn(z)=∑k=-nnak(z-c)k, then the sequence {Fn}{Fn} converges uniformly to ff on the compact set K.K.
  3. For any piecewise smooth geometric set S⊆Br(c),S⊆Br(c), whose boundary CSCS has finite length, and satisfying c∈S0,c∈S0, we have
    ∫CSf(ζ)dζ=2πia-1,∫CSf(ζ)dζ=2πia-1,
    (15)
    where a-1a-1 is the coefficient of (z-c)-1(z-c)-1 in the series of part (1).

Proof

Define numbers {ak}-∞∞{ak}-∞∞ as follows.

a k = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ . a k = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ .
(16)

Note that for any 0<δ<r0<δ<r we have from Cauchy's Theorem that

a k = 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) k + 1 d ζ , a k = 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) k + 1 d ζ ,
(17)

where CδCδ denotes the boundary of the disk B¯δ(c).B¯δ(c).

Let z≠cz≠c be in Br(c),Br(c), and choose δ>0δ>0 such that δ<|z-c|.δ<|z-c|. Then, using part (c) of (Reference), and then mimicking the proof of (Reference), we have

f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ - 1 2 π i ∫ C δ f ( ζ ) ζ - z d ζ = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) - ( z - c ) d ζ + 1 2 π i ∫ C δ f ( ζ ) ( z - c ) - ( ζ - c ) d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c 1 1 - z - c ζ - c d ζ + 1 2 π i ∫ C δ f ( ζ ) z - c 1 1 - ζ - c z - c d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c ∑ k = 0 ∞ ( z - c ζ - c ) k d ζ + 1 2 π i ∫ C δ f ( ζ ) z - c ∑ j = 0 ∞ ( ζ - c z - c ) j d ζ = ∑ k = 0 ∞ 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k + ∑ j = 0 ∞ 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) j d ζ ( z - c ) - j - 1 = ∑ k = 0 ∞ a k ( z - c ) k + ∑ k = - ∞ - 1 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = ∑ k = 0 ∞ a k ( z - c ) k + ∑ k = - ∞ - 1 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = ∑ k = - ∞ ∞ a k ( z - c ) k , f ( z ) = 1 2 π i ∫ C r f ( ζ ) ζ - z d ζ - 1 2 π i ∫ C δ f ( ζ ) ζ - z d ζ = 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) - ( z - c ) d ζ + 1 2 π i ∫ C δ f ( ζ ) ( z - c ) - ( ζ - c ) d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c 1 1 - z - c ζ - c d ζ + 1 2 π i ∫ C δ f ( ζ ) z - c 1 1 - ζ - c z - c d ζ = 1 2 π i ∫ C r f ( ζ ) ζ - c ∑ k = 0 ∞ ( z - c ζ - c ) k d ζ + 1 2 π i ∫ C δ f ( ζ ) z - c ∑ j = 0 ∞ ( ζ - c z - c ) j d ζ = ∑ k = 0 ∞ 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k + ∑ j = 0 ∞ 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) j d ζ ( z - c ) - j - 1 = ∑ k = 0 ∞ a k ( z - c ) k + ∑ k = - ∞ - 1 1 2 π i ∫ C δ f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = ∑ k = 0 ∞ a k ( z - c ) k + ∑ k = - ∞ - 1 1 2 π i ∫ C r f ( ζ ) ( ζ - c ) k + 1 d ζ ( z - c ) k = ∑ k = - ∞ ∞ a k ( z - c ) k ,
(18)

which proves part (1).

We leave the proofs of parts (2) and (3) to the exercises.

Exercise 6

  1. Justify bringing the summation signs out of the integrals in the calculation in the preceding proof.
  2. Prove parts (2) and (3) of the preceding theorem. Compare this with Exercise 4.

REMARK The representation of f(z)f(z) in the punctured disk Br'(c)Br'(c) given in part (1) of Theorem 3 and Theorem 4 is called the Laurent expansion of ff around the singularity c.c. Of course it differs from a Taylor series representation of f,f, as this one contains negative powers of z-c.z-c. In fact, which negative powers it contains indicates what kind of singularity the point cc is.

Non removable isolated singularities of a function ff share the property that the integral of ff around a disk centered at the singularity equals 2πia-1,2πia-1, where the number a-1a-1 is the coefficient of (z-c)-1(z-c)-1 in the Laurent expansion of ff around c.c. This number 2πia-12πia-1 is obviously significant, and we call it the residue of f at c, and denote it by Rf(c).Rf(c).

Combining Theorem 1, Theorem 3, and Theorem 4, we obtain:

Theorem 5: Residue Theorem

Let SS be a piecewise smooth geometric set whose boundary has finite length, let c1,...,cnc1,...,cn be points in S0,S0, and suppose ff is a complex-valued function that is continuous at every point zz in SS except the ckck's, and differentiable at every point z∈S0z∈S0 except at the ckck's. Assume finally that each ckck is a nonremovable isolated singularity of f.f. Then

∫ C S f ( ζ ) d ζ = ∑ k = 1 n R f ( c k ) . ∫ C S f ( ζ ) d ζ = ∑ k = 1 n R f ( c k ) .
(19)

That is, the contour integral around CSCS is just the sum of the residues inside S.S.

Exercise 7

Prove Theorem 5.

Exercise 8

Use the Residue Theorem to compute ∫CSf(ζ)dζ∫CSf(ζ)dζ for the functions ff and geometric sets SS given below. That is, determine the poles of ff inside S,S, their orders, the corresponding residues, and then evaluate the integrals.

  1. f(z)=sin(3z)/z2,f(z)=sin(3z)/z2, and S=B¯1(0).S=B¯1(0).
  2. f(z)=e1/z,f(z)=e1/z, and S=B¯1(0).S=B¯1(0).
  3. f(z)=e1/z2,f(z)=e1/z2, and S=B¯1(0).S=B¯1(0).
  4. f(z)=(1/z(z-1)),f(z)=(1/z(z-1)), and S=B¯2(0).S=B¯2(0).
  5. f(z)=((1-z2)/z(1+z2)(2z+1)2),f(z)=((1-z2)/z(1+z2)(2z+1)2), and S=B¯2(0).S=B¯2(0).
  6. f(z)=1/(1+z4)=(1/(z2-i)(z2+i)),f(z)=1/(1+z4)=(1/(z2-i)(z2+i)), and S=B¯r(0)S=B¯r(0) for any r>1.r>1.

The Residue Theorem, a result about contour integrals of functions of a complex variable, can often provide a tool for evaluating integrals of functions of a real variable.

Example 1

Consider the integral

∫ - ∞ ∞ 1 1 + x 4 d x . ∫ - ∞ ∞ 1 1 + x 4 d x .
(20)

Let us use the Residue Theorem to compute this integral.

Of course what we need to compute is

lim B → ∞ ∫ - B B 1 1 + x 4 d x . lim B → ∞ ∫ - B B 1 1 + x 4 d x .
(21)

The first thing we do is to replace the real variable xx by a complex variable Z,Z, and observe that the function f(z)=1/(1+z4)f(z)=1/(1+z4) is analytic everywhere except at the four points ±eiπ/4±eiπ/4 and ±e3iπ/4.±e3iπ/4. See part (f) of the preceding exercise. These are the four points whose fourth power is -1,-1, and hence are the poles of the function f.f.

Next, given a positive number B,B, we consider the geometric set (rectangle) SBSB that is determined by the interval [-B,B][-B,B] and the two bounding functions l(x)=0l(x)=0 and u(x)=B.u(x)=B. Then, as long as B>1,B>1, we know that ff is analytic everywhere in S0S0 except at the two points c1=eiπ/4c1=eiπ/4 and c2=e3iπ/4,c2=e3iπ/4, so that the contour integral of ff around the boundary of SBSB is given by

∫ C S B 1 1 + ζ 4 d ζ = R f ( c 1 ) + R f ( c 2 ) . ∫ C S B 1 1 + ζ 4 d ζ = R f ( c 1 ) + R f ( c 2 ) .
(22)

Now, this contour integral consists of four parts, the line integrals along the bottom, the two sides, and the top. The magic here is that the integrals along the sides, and the integral along the top, all tend to 0 as BB tends to infinity, so that the integral along the bottom, which after all is what we originally were interested in, is in the limit just the sum of the residues inside the geometric set.

Exercise 9

Verify the details of the preceding example.

  1. Show that
    limB→∞∫0B11+(B+it)4dt=0.limB→∞∫0B11+(B+it)4dt=0.
    (23)
  2. Verify that
    limB→∞∫-BB11+(t+iB)4dt=0.limB→∞∫-BB11+(t+iB)4dt=0.
    (24)
  3. Show that
    ∫-∞∞11+x4dx=π2.∫-∞∞11+x4dx=π2.
    (25)

Methods similar to that employed in the previous example and exercise often suffice to compute integrals of real-valued functions. However, the method may have to be varied. For instance, sometimes the appropriate geometric set is a rectangle below the xx-axis instead of above it, sometimes it should be a semicircle instead of a rectangle, etc. Indeed, the choice of contour (geometric set) can be quite subtle. The following exercise may shed some light.

Exercise 10

  1. Compute
    ∫-∞∞eix1+x4dx∫-∞∞eix1+x4dx
    (26)
    and
    ∫-∞∞e-ix1+x4dx.∫-∞∞e-ix1+x4dx.
    (27)
  2. Compute
    ∫-∞∞sin(-x)1+x3dx∫-∞∞sin(-x)1+x3dx
    (28)
    and
    ∫-∞∞sinx1+x3dx.∫-∞∞sinx1+x3dx.
    (29)

Example 2

An historically famous integral in analysis is ∫-∞∞sinx/xdx.∫-∞∞sinx/xdx. The techniques described above don't immediately apply to this function, for, even replacing the xx by a z,z, this function has no poles, so that the Residue Theorem wouldn't seem to be much help. Though the point 0 is a singularity, it is a removable one, so that this function sinz/zsinz/z is essentially analytic everywhere in the complex plane. However, even in a case like this we can obtain information about integrals of real-valued functions from theorems about integrals of complex-valued functions.

Notice first that ∫-∞∞sinx/xdx∫-∞∞sinx/xdx is the imaginary part of ∫-∞∞eix/xdx,∫-∞∞eix/xdx, so that we may as well evaluate the integral of this function. Let ff be the function defined by f(z)=eiz/z,f(z)=eiz/z, and note that 0 is a pole of order 1 of f,f, and that the residue Rf(0)=2πi.Rf(0)=2πi. Now, for each B>0B>0 and δ>0δ>0 define a geometric set SB,δ,SB,δ, determined by the interval [-B,B],[-B,B], as follows: The upper bounding function uB,δuB,δ is given by uB,δ(x)=B,uB,δ(x)=B, and the lower bounding function lB,δlB,δ is given by lB,δ(x)=0lB,δ(x)=0 for -B≤x≤-δ-B≤x≤-δ and δ≤x≤B,δ≤x≤B, and lB,δ(x)=δeiπx/δlB,δ(x)=δeiπx/δ for -δ<x<δ.-δ<x<δ. That is, SB,δSB,δ is just like the rectangle SBSB in Example 1 above, except that the lower boundary is not a straight line. Rather, the lower boundary is a straight line from -B-B to -δ,-δ, a semicircle below the xx-axis of radius δδ from -δ-δ to δ,δ, and a straight line again from δδ to B.B.

By the Residue Theorem, the contour integral

∫ C S B , δ f ( ζ ) d ζ = R f ( 0 ) = 2 π i . ∫ C S B , δ f ( ζ ) d ζ = R f ( 0 ) = 2 π i .
(30)

As in the previous example, the contour integrals along the two sides and across the top of SB,δSB,δ tend to 0 as BB tends to infinity. Finally, according to part (e) of (Reference), the contour integral of ff along the semicircle in the lower boundary is πiπi independent of the value of δ.δ. So,

lim B → ∞ lim δ → 0 ∫ graph ( l B , δ ) e i ζ ζ d ζ = π i , lim B → ∞ lim δ → 0 ∫ graph ( l B , δ ) e i ζ ζ d ζ = π i ,
(31)

implying then that

∫ - ∞ ∞ sin x x d x = π . ∫ - ∞ ∞ sin x x d x = π .
(32)

Exercise 11

  1. Justify the steps in the preceding example. In particular, verify that
    limB→∞∫0Bei(B+it)B+itdt=0,limB→∞∫0Bei(B+it)B+itdt=0,
    (33)
    limB→∞∫-BBei(t+iB)t+iBdt=0,limB→∞∫-BBei(t+iB)t+iBdt=0,
    (34)
    and
    ∫Cδeiζζdζ=πi,∫Cδeiζζdζ=πi,
    (35)
    where CδCδ is the semicircle of radius δ,δ, centered at the origin and lying below the xx-axis.
  2. Evaluate
    ∫-∞∞sin2xx2dx.∫-∞∞sin2xx2dx.
    (36)

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